Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 213022 by Spillover last updated on 28/Oct/24

	Answered by TonyCWX08 last updated on 31/Oct/24

	$F C = a + b$ $h e i g h t o f {t r a p e z i u m}_{A F B C}$ $= \sqrt{b^{2} - {(\frac{b}{2})}^{2}}$ $= \sqrt{\frac{3 b^{2}}{4}}$ $= \frac{b \sqrt{3}}{2}$ ${A r e a}_{A F B C}$ $= \frac{1}{2} (a + a + b) (\frac{b \sqrt{3}}{2})$ $= \frac{2 a b \sqrt{3} + b^{2} \sqrt{3}}{4}$ $= \frac{a b \sqrt{3}}{2} + {(\frac{b}{2})}^{2} \sqrt{3}$ $= \sqrt{3} (\frac{a b}{2} + (\frac{b^{2}}{2}))$ $L i t t l e e q u i l a t e r a l t r i a n g l e f o r m e d w h e n j o i n i n g p o i n t A t o D a n d F t o C$ $S i d e l e n g t h = b$ $A r e a = \frac{b^{2}}{4} \sqrt{3}$ $B i g e q u i l a t e r a l t r i a n g l e f o r m e d i n t h e m i d d l e o f t h e H e x a g o n w h e n j o i n i n g t h e s a m e p o i n t s$ $S i d e l e n g t h$ $= a + b - 2 b$ $= a - b$ $A r e a = \frac{{(a - b)}^{2}}{4} \sqrt{3}$ $A r e a o f C y c l i c H e x a g o n$ $= 3 {A r e a}_{A F B C} - 3 {A r e a}_{S m a l l Δ} + {A r e a}_{B i g Δ}$ $= 3 (\sqrt{3} (\frac{a b}{2} + \frac{b^{2}}{4})) - 3 (\frac{b^{2}}{4} \sqrt{3}) + \frac{a^{2} - 2 a b + b^{2}}{4} \sqrt{3}$ $= \frac{3 a b \sqrt{3}}{2} + \frac{3 \sqrt{3} b^{2}}{4} - \frac{3 b^{2} \sqrt{3}}{4} + \frac{\sqrt{3} (a^{2} - 2 a b + b^{2})}{4}$ $= \frac{6 a b \sqrt{3} + 3 b^{2} \sqrt{3} - 3 b^{2} \sqrt{3} + a^{2} \sqrt{3} - 2 a b \sqrt{3} + b^{2} \sqrt{3}}{4}$ $= \frac{a^{2} \sqrt{3} + 4 a b \sqrt{3} + b^{2} \sqrt{3}}{4}$ $= \sqrt{3} [\frac{a^{2}}{4} + a b + \frac{b^{2}}{4}]$ $= \sqrt{3} [{(\frac{a}{2})}^{2} + {(\frac{b}{2})}^{2} + a b]$ $H e n c e P r o v e d .$
	Answered by TonyCWX08 last updated on 31/Oct/24

	${A r e a}_{P Q R} = \frac{X^{2}}{4} \sqrt{3}$ $\frac{X^{2}}{4} \sqrt{3} = \sqrt{3} [{(\frac{b}{2})}^{2} + {(\frac{a}{2})}^{2} + a b]$ ${(\frac{X}{2})}^{2} = [{(\frac{a}{2})}^{2} + a b + {(\frac{b}{2})}^{2}]$ $\frac{X}{2} = \sqrt{{(\frac{a}{2})}^{2} + a b + {(\frac{b}{2})}^{2}}$ $\frac{X}{2} = \sqrt{\frac{a^{2}}{4} + \frac{4 a b}{4} + \frac{b^{2}}{4}}$ $\frac{X}{2} = \sqrt{\frac{a^{2} + 4 a b + b^{2}}{4}}$ $\frac{X}{2} = \frac{\sqrt{a^{2} + 4 a b + b^{2}}}{2}$ $X = \sqrt{a^{2} + 4 a b + b^{2}}$
	Answered by TonyCWX08 last updated on 31/Oct/24

	Commented by Spillover last updated on 31/Oct/24

	$g r e a t w o r k t h a n k y o u$
	Commented by TonyCWX08 last updated on 01/Nov/24

	$W e l c o m e .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum